Answer
$n=1414$
Work Step by Step
$level~of~confiance=(1-α).100$%
$94$% $=(1-α).100$%
$0.94=1-α$
$α=0.06$
$z_{\frac{α}{2}}=z_{0.03}$
If the area of the standard normal curve to the right of $z_{0.03}$ is 0.03, then the area of the standard normal curve to the left of $z_{0.03}$ is $1−0.03=0.97$
According to Table V, the z-score which gives the closest value to 0.97 is 1.88.
Now, the sample size (no prior estimate of $p ̂$) :
$E=0.025$ (within 2.5 percentage points)
$n=0.25(\frac{z_{\frac{α}{2}}}{E})^2$
$n=0.25(\frac{1.88}{0.025})^2$
$n=1413.76$
Round up:
$n=1414$
